How do you prove #Sin^2x-sin^2y=sin(x+y)sin(x-y)#?

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Answer 1 · 2016-05-24T06:19:12

Please see below.

#sin(x+y)sin(x-y)#

= #(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)#

= #sin^2xcos^2y-cos^2xsin^2y#

= #sin^2x(1-sin^2y)-(1-sin^2x)sin^2y#

= #sin^2x-sin^2xsin^2y-sin^2y+sin^2xsin^2y#

= #sin^2x-cancel(sin^2xsin^2y)-sin^2y+cancel(sin^2xsin^2y)#

= #sin^2x-sin^2y#